linearization and differentials: pr\n(1 point)\nfind the differential of ( y=sqrt{10 + t^{2}} ).\n( d…

Explanation:

Step1: Rewrite the function

Rewrite (y = \sqrt{10 + t^{2}}=(10 + t^{2})^{\frac{1}{2}}).

Step2: Apply the chain - rule

The chain - rule states that if (y = f(g(t))), then (y^\prime=f^\prime(g(t))\cdot g^\prime(t)). Let (u = 10 + t^{2}), so (y = u^{\frac{1}{2}}). First, find (\frac{dy}{du}): (\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}). Second, find (\frac{du}{dt}): (\frac{du}{dt}=2t). Then (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}). Substitute (u = 10 + t^{2}) into (\frac{dy}{du}), we get (\frac{dy}{dt}=\frac{1}{2\sqrt{10 + t^{2}}}\cdot2t).

Step3: Simplify the expression

(\frac{dy}{dt}=\frac{t}{\sqrt{10 + t^{2}}}). Since (dy=\frac{dy}{dt}dt), so (dy=\frac{t}{\sqrt{10 + t^{2}}}dt).

Answer:

(\frac{t}{\sqrt{10 + t^{2}}})